Linear Algebra: Linear Equation System -> Parameter Form

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The discussion revolves around solving a linear system of equations using the Gaussian elimination method and expressing the solution set in parameter form. After applying Gaussian elimination, the reduced row echelon form indicates a rank of 3, allowing for one free variable, which is chosen as x4. The derived equations lead to a plane equation expressed as x1 + x2 + 2x3 - 5x4 - 2.5 = 0, with the normal vector identified as (1, 1, 2, -5). Participants discuss how to derive parameter forms from the plane equation by finding vectors perpendicular to the normal vector. The conversation emphasizes the importance of understanding the transition from standard equation forms to parameter forms in linear algebra.
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Homework Statement


Consider the following linear system of equations:
x1+2x3-5x4 = 0
x1 + 4x2 +4x3 – 5x4 = 10
x1 + 2x2 + 3x3 – 5x4 = 5
4x1 + 2x2 + 9x3 – 20x4 = 5

b) Solve the equation system with the Gaussian method.
c) The solution set describes a plane. Specify it in the parameter form.

Homework Equations


Parameter form: \vec{r}= \vec{r}0 + λ\vec{v} + β\vec{w}

The Attempt at a Solution


3. My Work
After Gauss?:
1 2 0 -5 0
0 2 4 0 10
0 0 3 0 5
0 0 0 0 0

Rank is 3 so there is one free chooseable variable. I chose x4 and ended with the equation:
\vec{X} =
\begin{vmatrix}-10/3\\5/3\\5/3\\0\end{vmatrix} + \vec{x}4 * \begin{vmatrix}5\\0\\0\\1\end{vmatrix}

Which I think is the plane’s formula? I’m not totally sure. My teacher said that he wouldn’t be putting the answers up, so I wanted a second opinion on my numbers. Please and thank you.
 
Last edited:
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Hey daemon_dkm and welcome to the forums.

Plugging this into octave, I got the following reduced row echelon form:

1.00000 0.00000 2.00000 -5.00000 0.00000
0.00000 1.00000 0.50000 0.00000 2.50000
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000

This means setting x4 as our dummy we get x2 + x3 = 2.5 and x1 + x3 = 5x4 which means that adding the two together gives us x2 + x3 + x1 + x3 = 2.5 + 5x4 then minus x4 to LHS gives the plane equation:

x1 + x2 + 2x3 - 5x4 - 2.5 = 0, which can be converted to n . (r - r0) = 0 where n is the normal vector, r is an arbitrary point and r0 is a vector one the plane. the normal is anything that is a multiple (except 0) of the vector (1,1,2,-5) and n.r0 = 2.5. where you can solve for any value of r0 you wish.

Do you know how to go from ax + by + cz + dw + e to other formulas?
 
ax + by + cz + dw + e

the normal vector from that plane equation is n= (a, b, c, d), so then you need to find a vector that's perpendicular to the normal vector. Then you need to find another vector that's perpendicular to both of them to use them for the parameter form?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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